Question: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $r \neq 0$. $q = \dfrac{r^2 - 81}{-10r + 90} \div \dfrac{-2r - 18}{r + 8} $
Explanation: Dividing by an expression is the same as multiplying by its inverse. $q = \dfrac{r^2 - 81}{-10r + 90} \times \dfrac{r + 8}{-2r - 18} $ First factor the quadratic. $q = \dfrac{(r + 9)(r - 9)}{-10r + 90} \times \dfrac{r + 8}{-2r - 18} $ Then factor out any other terms. $q = \dfrac{(r + 9)(r - 9)}{-10(r - 9)} \times \dfrac{r + 8}{-2(r + 9)} $ Then multiply the two numerators and multiply the two denominators. $q = \dfrac{ (r + 9)(r - 9) \times (r + 8) } { -10(r - 9) \times -2(r + 9) } $ $q = \dfrac{ (r + 9)(r - 9)(r + 8)}{ 20(r - 9)(r + 9)} $ Notice that $(r - 9)$ and $(r + 9)$ appear in both the numerator and denominator so we can cancel them. $q = \dfrac{ \cancel{(r + 9)}(r - 9)(r + 8)}{ 20(r - 9)\cancel{(r + 9)}} $ We are dividing by $r + 9$ , so $r + 9 \neq 0$ Therefore, $r \neq -9$ $q = \dfrac{ \cancel{(r + 9)}\cancel{(r - 9)}(r + 8)}{ 20\cancel{(r - 9)}\cancel{(r + 9)}} $ We are dividing by $r - 9$ , so $r - 9 \neq 0$ Therefore, $r \neq 9$ $q = \dfrac{r + 8}{20} ; \space r \neq -9 ; \space r \neq 9 $